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Electrochemistry question

2009 · Q97
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Electrochemistry question

2009 · Q97

NEETChemistryElectrochemistryMCQ+4 / −1
Given :
(i)   Cu2+ + 2e−-− →\to→ Cu,  Eo = 0.337 V
(ii)  Cu2+ + e−-− →\to→ Cu+,  Eo = 0.153 V
Electrode potential, Eo for the reaction,
Cu+ + e−-− →\to→ Cu,  will be
  1. A
    0.90 V
  2. B
    0.30 V
  3. C
    0.38 V
  4. D
    0.52 V
View written solutionFree

Correct answer: D

For the reaction,

Cu2+ + 2e−-− →\to→ Cu,  Eo = 0.337 V

Δ\Delta ΔGo = - nFEo

= – 2 × F × 0.337

= – 0.674 F ......(i)

For the reaction,

Cu2+ + e−-− →\to→ Cu+,  Eo = 0.153 V

Δ\Delta ΔGo = - nFEo

= – 1 × F × – 0.153

= 0.153 F

On adding eqn (i) & (ii)

Cu2+ + e−-− →\to→ Cu+

Δ\Delta ΔGo = –0.521 F = –nFE°

⇒\Rightarrow⇒ E° = 0.52 V

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